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3 years ago

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two barrels contain equal quantities of honey from one barrel 37 gallons of Honey or drone from the other Barrel 7 gallons of Ho

ney are drawn the quantity remaining in Winnsboro is now seven times that remaining in the other Barrel how much did each Barrel contain at first

2 answers:

AlexFokin [52]3 years ago

6 0

Answer: At the start, each barrel has the same amount of honey, so:

Barrel 1 has X gallons

Barrel 2 has X gallons.

from 1, they extract 37 gallos, and from 2 they extract 7 gallons, so now:

Barrel 1 has X - 37

Barrel 2 has X- 7

and now, the quantity in the Barrel 2 is seven times the quantity in the barrel 1, so:

X- 7 = 7*(X - 37)

we need to find X.

X - 7 = 7*X - 37*7 = 7*X - 259

X - 7*X = 7 - 259

-6*X = -252

X = 252/6 = 42

So at the start, each barrel has 42 gallons of honey.

Elena L [17]3 years ago

5 0

308 gallons 37 plus 7 times7

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Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $

Solnce55 [7]

Answer:

a) $\hat a = max(X_i)$

For this case the value for $\hat a$ is always smaller than the value of a, assuming $X_i \sim Unif[0,a]$ So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

$E(a) - a= 0$ and that's not our case.

b) $E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

c) $P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

$f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]$

e) On this case we see that the estimator $\hat a_1$ is better than $\hat a_2$ and the reason why is because:

$V(\hat a_1) > V(\hat a_2)$

$\frac{a^2}{3n}> \frac{a^2}{n(n+2)}$

$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming $X_1, X_2 , ..., X_n \sim U(0,a)$

And we are are ssuming the following estimator:

$\hat a = max(X_i)$

For this case the value for $\hat a$ is always smaller than the value of a, assuming $X_i \sim Unif[0,a]$ So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

$E(a) - a= 0$ and that's not our case.

Part b

For this case we assume that the estimator is given by:

$E(\hat a) = \frac{na}{n+1}$

And using the definition of bias we have this:

$E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

Part c

For this case we the followng random variable $Y = max (X_i)$ and we can find the cumulative distribution function like this:

$P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

Since all the random variables have the same distribution.

Now we can find the density function derivating the distribution function like this:

$f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]$

Now we can find the expected value for the random variable Y and we got this:

$E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}$

And the bias is given by:

$E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}$

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

$V(\hat a_1) = \frac{a^2}{3n}$

$V(\hat a_2) = \frac{a^2}{n(n+2)}$

On this case we see that the estimator $\hat a_1$ is better than $\hat a_2$ and the reason why is because:

$V(\hat a_1) > V(\hat a_2)$

$\frac{a^2}{3n}> \frac{a^2}{n(n+2)}$

$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

8 0

4 years ago

Guys do you know this?

TEA [102]

Answer:

That would be 2/7

Step-by-step explanation:

Using substitution. x^2 +x+1 at 1 is 7

sqrt4 is 2

2/7

6 0

3 years ago

Geometry help please!!!

julia-pushkina [17]

Before we begin, let's identify what kind of angles these are and are they related in any way?

These angles are both acute and they are both corresponding angles.
Corresponding angles are equal to each other, and we can use this fact to our advantage.

Since they are equal to each other, we can set the equations of 1 and 2 equal to each other. Like so,

1 = 2
83 - 2x = 92 - 3x

Now, we can solve for X by isolating it on one side.

83 - 2x = 92 - 3x

Add 3x to each side: (This basically moves the X on the right side to the left.)
83 - 2x + 3x = 92 - 3x + 3x
83 + x = 92

Subtract 83 on each side to isolate the X.
83 + x - 83 = 92 - 83
x = 92 - 83
x = 9

Therefore, X equals 9. To check our work, we can substitute X for 9.
83 - 2(9) = 92 - 3(9)
83 - 18 = 92 - 27
65 = 65 -
TRUE

So to conclude, Angle 1 is 65 degrees, Angle 2 is 65 degrees, and X equals 9.

Hope I could help you out!
If my answer is incorrect, or I provided an answer you were not looking for, please let me know. However, if my answer is explained well and correct, please consider marking my answer as Brainliest! :)

Have a good one.
God bless!

7 0

3 years ago

I need help with this question it’s urgentttt

Andreas93 [3]

Answer:

EF is equal to 6.

Step-by-step explanation:

Those triangles are similar so put them in proportion , as in 8/16=EF/12 then cross multiply and solve

4 0

3 years ago

What is the hydronium ion (H3O+) concentration in milk if it has a pH of 8.2?

valkas [14]

The hydronium ion is H₃O⁺ and its concentration is equivalent to the H⁺ ion concentration. The H⁺ ion concentration is given by:

pH = -log[H⁺]
[H⁺] = 10^(-8.2)
[H⁺] = 6.31 x 10⁻⁹ mole/dm³

The hydronium ion concentration is 6.31 x 10⁻⁹ mole/dm³

5 0

4 years ago